integral from 1 to 4 of (u-2)/(sqrt(u))

解

\int_{1}^{4} \frac{u - 2}{u} d u

\frac{2}{3}

十進法表記

0.66666 \dots

解答ステップ

\int_{1}^{4} \frac{u - 2}{u} d u

拡張

\frac{u - 2}{u} : u - \frac{2}{u}

= \int_{1}^{4} u - \frac{2}{u} d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{4} u d u - \int_{1}^{4} \frac{2}{u} d u

\int_{1}^{4} u d u = \frac{14}{3}

\int_{1}^{4} \frac{2}{u} d u = 4

= \frac{14}{3} - 4

簡素化

\frac{14}{3} - 4 : \frac{2}{3}

= \frac{2}{3}